chapter 12 temperature and heat. liquid water (hot) water vapor (hot) frozen water (snow) (cold)...
TRANSCRIPT
liquid water (hot)liquid water (hot)water vapor (hot)water vapor (hot)
frozen water (snow) (cold)frozen water (snow) (cold)
monkey (warm)monkey (warm)
•A system composed by many parts at different temperatures is not in “thermal equilibrium. •Heat flows from hot parts to cold parts. Cold parts heat up, hot parts cool down.• Some energy is spent into evaporating water and melting snow.
•A system composed by many parts at different temperatures is not in “thermal equilibrium. •Heat flows from hot parts to cold parts. Cold parts heat up, hot parts cool down.• Some energy is spent into evaporating water and melting snow.
12.1 Temperature
Temperature measures the kinetic energy of molecules. Temperature measures the kinetic energy of molecules.
hot water, molecules moving fast
freezing water, molecules move slow
12.1 Common Temperature Scales
Temperatures are reported in degreesCelsius or degrees Fahrenheit.
Temperatures changed, on theother hand, are reported in Celsius degrees or Fahrenheit degrees:
F 5
9C 1
(TF −32)×59 =TC
conversion F to C:
conversion C to F
TC ×95 + 32 =TF
12.1 Common Temperature Scales
Example 1 Converting from a Fahrenheit to a Celsius Temperature
A healthy person has an oral temperature of 98.6oF. What would thisreading be on the Celsius scale?
(98.6 −32)×59 =37 Co
T in F T in C
(TF −32)×59 =TC
12.1 Common Temperature Scales
Example 2 Converting from a Celsius to a Fahrenheit Temperature
The temperature in a cold day in Alaska is -20.0oC. Find the corresponding temperature on the Fahrenheit scale.
−20 × 95 + 32 = −4 Fo
T in C T in F
TC ×95 + 32 =TF
12.1 Common Temperature Scales
Example 2 Converting from a Celsius to a Fahrenheit Temperature
The temperature in the classroom is +20.0oC. Find the corresponding temperature onthe Fahrenheit scale.
20 ×95 + 32 =68 Fo
T in C T in F
12.2 The Kelvin Temperature Scale
15.273 cTT
Kelvin temperature
Minimum possible in the entire universe!
Temperature measures the kinetic energy of molecules.
zero Kelvin means zero kinetic energy. Can’t be lower than that!
Minimum possible in the entire universe!
Temperature measures the kinetic energy of molecules.
zero Kelvin means zero kinetic energy. Can’t be lower than that!
12.4 Linear Thermal Expansion
LINEAR THERMAL EXPANSION OF A SOLID
The length of an object changes when its temperature changes:
TLL o
coefficient of linear expansion
Common Unit for the Coefficient of Linear Expansion: 1C
C
1
12.4 Linear Thermal Expansion
Example 3 The Buckling of a Sidewalk
A concrete sidewalk is constructed between two buildings on a day when the temperature is 25oC. As the temperature rises to 38oC, the slabs expand, but no space is provided forthermal expansion. Determine the distance yin part (b) of the drawing.
12.4 Linear Thermal Expansion
L = α LoΔT
= 12 ×10−6 Co( )−1⎡
⎣⎤⎦
3.0 m( ) 13 Co( ) = 0.00047 m
m 053.0m 00000.3m 00047.3 22 y
12.4 Linear Thermal Expansion
THE EXPANSION OF HOLES
Conceptual Example 5 The Expansion of Holes
The figure shows eight square tiles that are arranged to form a square patternwith a hold in the center. If the tiled are heated, what happens to the size of the hole?
12.4 Linear Thermal Expansion
A hole in a piece of solid material expands when heated and contracts whencooled, just as if it were filled with the material that surrounds it.
12.5 Volume Thermal Expansion
VOLUME THERMAL EXPANSION
The volume of an object changes when its temperature changes:
TVV o
coefficient of volume expansion
Common Unit for the Coefficient of Volume Expansion: 1C
C
1
12.5 Volume Thermal Expansion
Example 8 An Automobile Radiator
A small plastic container, called the coolant reservoir, catchesthe radiator fluid that overflows when an automobile enginebecomes hot. The radiator is made of copper and the coolant has an expansion coefficient of 4.0x10-4 (Co)-1. If the radiator is filled to its 15-quart capacitywhen the engine is cold (6oC),how much overflow will spill into the reservoir when the coolant reaches its operating temperature (92oC)?
12.5 Volume Thermal Expansion
Vcoolant = 4.10 × 10−4 Co
( )−1
( ) 15 quarts( ) 86 Co( ) = 0.53 quarts
Vradiator = 51 × 10−6 Co
( )−1
( ) 15 quarts( ) 86 Co( ) = 0.066 quarts
quarts 0.46quarts 066.0quarts 53.0spill V
12.6 Heat and Internal Energy
DEFINITION OF HEAT
Heat is a flow of internal energy between two objects
Heat spontaneously flows from a higher-temperature object to a lower-temperature object because of a difference in temperatures.
When the temperature difference becomes zero the objects are in “thermal equilibrium” and the heat flow stops.
Heat can also flow from a cold object to a hot object, although the process is not spontaneous and requires external work (air conditioner)
SI Unit of Heat: joule (J)
12.6 Heat and Internal Energy
The heat that flows from hot to cold originates in the internal energy ofthe hot substance.
It is not correct to say that a substancecontains heat.
12.7 Heat and Temperature Change: Specific Heat Capacity
SOLIDS AND LIQUIDS
HEAT SUPPLIED OR REMOVED IN CHANGING THE TEMPERATUREOF A SUBSTANCE
The heat that must be supplied or removed to change the temperature ofa substance is
TmcQ
specific heatcapacity
Common Unit for Specific Heat Capacity: J/(kg·Co)
12.7 Heat and Temperature Change: Specific Heat Capacity
Example 9 A Hot Jogger
In a half-hour, a 65-kg jogger can generate 8.0x105J of heat. This heatis removed from the body by a variety of means, including the body’s owntemperature-regulating mechanisms. If the heat were not removed, how much would the body temperature increase?
TmcQ
T =Q
mc=
8.0 ×105 J
65 kg( ) 3500 J kg ⋅Co( )⎡⎣ ⎤⎦= 3.5 Co
12.7 Heat and Temperature Change: Specific Heat Capacity
GASES
The value of the specific heat of a gas depends on whether the pressure orvolume is held constant.
This distinction is not important for solids.
OTHER UNITS
1 kcal = 4186 joules
1 cal = 4.186 joules
12.7 Heat and Temperature Change: Specific Heat Capacity
CALORIMETRY
If there is no heat loss to the surroundings,the heat lost by the hotter object equals theheat gained by the cooler ones.
12.7 Heat and Temperature Change: Specific Heat Capacity
Example 12 Measuring the Specific Heat Capacity
The calorimeter is made of 0.15 kg of aliminumand contains 0.20 kg of water. Initially, thewater and cup have the same temperatureof 18.0oC. A 0.040 kg mass of unknown material is heated to a temperature of 97.0oC and then added to the water.
After thermal equilibrium is reached, thetemperature of the water, the cup, and the material is 22.0oC. Ignoring the small amountof heat gained by the thermometer, find the specific heat capacity of theunknown material.
12.7 Heat and Temperature Change: Specific Heat Capacity
unknownwaterAl TmcTmcTmc
cunknown =mcT( )Al
+ mcT( )water
mT( )unknown
=9.00 ×102 J kg⋅Co( )⎡⎣ ⎤⎦ 0.15 kg( ) 4.0 Co( ) + 4186 J kg⋅Co( )⎡⎣ ⎤⎦ 0.20 kg( ) 4.0 Co( )
0.040 kg( ) 75.0 Co( )
=1300 J kg⋅Co( )
12.8 Heat and Phase Change: Latent Heat
During a phase change, the temperature of the mixture does not change (provided the system is in thermal equilibrium).
12.8 Heat and Phase Change: Latent Heat
Conceptual Example 13 Saving Energy
Suppose you are cooking spaghetti for dinner, and the instructionssay “boil pasta in water for 10 minutes.” To cook spaghetti in an openpot with the least amount of energy, should you turn up the burnerto its fullest so the water vigorously boils, or should you turn downthe burner so the water barely boils?
12.8 Heat and Phase Change: Latent Heat
HEAT SUPPLIED OR REMOVED IN CHANGING THE PHASEOF A SUBSTANCE
The heat that must be supplied or removed to change the phaseof a mass m of a substance is
mLQ
latent heat
SI Units of Latent Heat: J/kg
12.8 Heat and Phase Change: Latent Heat
Example 14 Ice-cold Lemonade
Ice at 0oC is placed in a Styrofoam cup containing 0.32 kg of lemonadeat 27oC. The specific heat capacity of lemonade is virtually the same asthat of water. After the ice and lemonade reach and equilibriumtemperature, some ice still remains. Assume that mass of the cup isso small that it absorbs a negligible amount of heat.
lemonade
bylost Heat
lemonade
iceby gainedHeat
iceTcmmL f
12.8 Heat and Phase Change: Latent Heat
lemonade
bylost Heat
lemonade
iceby gainedHeat
iceTcmmL f
mice cmT lemonade
Lf
4186J kg⋅Co ⎡⎣ ⎤⎦ 0.32 kg 27oC−0oC
3.35 ×105 J kg0.11 kg
12.9 Equilibrium Between Phases of Matter
The pressure of vapor that coexists in equilibrium with the liquid iscalled the equilibrium vapor pressure of the liquid.
12.9 Equilibrium Between Phases of Matter
Only when the temperature and vapor pressure correspond to a pointon the curved line do the liquid and vapor phases coexist in equilibrium.
liquid
vapor
12.9 Equilibrium Between Phases of Matter
As is the case for liquid/vapor equilibrium, a solid can be inequilibrium with its liquid phaseonly at specific conditions oftemperature and pressure.
12.10 Humidity
Air is a mixture of gases.
The total pressure is the sum of the partial pressures of the componentgases.
The partial pressure of water vapor depends on weather conditions. It can be as low as zero or as high as the vapor pressure of water at thegiven temperature.
100
re temperatuexistingat water of pressure vapor mEquilibriu
or water vapof pressure Partialhumidity relativePercent
To provide an indication of how much water vapor is in the air, weatherforecasters usually give the relative humidity:
12.10 Humidity
Example 17 Relative Humidities
One day, the partial pressure of water vapor is 2.0x103 Pa. Using thevaporization curve, determine the relative humidity if the temperatureis 32oC.
12.10 Humidity
100
re temperatuexistingat water of pressure vapor mEquilibriu
or water vapof pressure Partialhumidity relativePercent
%42100Pa 108.4
Pa 100.2humidity Relative
3
3